Web: http://www.maths.otago.ac.nz/jmc
Twitter: @uojmc
First
Year 9 (Form 3) Prize Winners
Jun Min Hong
Tauranga Boys’ College
Second
Claire Shu
Avondale College
Third
Alphaeus Ang
Liston College
Top 30 (in School Order):
Evan Sun, Auckland Grammar School
Jaqlin van Schalkwyk, Kristin School
Songyan Teng, Pakuranga College
Mukund Karthik, Botany Downs Secondary College
Kevin Guan, Macleans College
Tide Sun, Pinehurst School
Jing Qu, Botany Downs Secondary College
William Han, Macleans College
Eric Li, St Kentigern College
Yash Shahri, Botany Downs Secondary College
Kevin Hou, Macleans College
Lucy Nie, St Kentigern College
Daniel Chong, Christchurch Boys’ High School
Richard Jiang, Macleans College
Neil Stephen Parinas, Tauhara College
Patrick Zhang, John McGlashan College
Sam Shi, Macleans College
Ben (Bowoo) Kang, Tauranga Boys’ College
Isaac Chandler, Kerikeri High School
Angela Wang, Macleans College
Sam Kang, Tauranga Boys’ College
Marcus Ooi, King’s College
Michael Wang, Macleans College
William Price, Westlake Boys’ High School
William Scharpf, King’s High School
Matthew Creahan, Pakuranga College
David Yao, Westlake Boys’ High School
Year 10 (Form 4) Prize Winners
First
Xutong Wang
Auckland Grammar School
Second
Chuanye Chen
Auckland Grammar School
Third
Boen Deng
John McGlashan College
Top 30 (in School Order):
Max Gu, Auckland Grammar School
Wills Wynn Thomas, Christ’s College
Ewan Smith, Otamatea High School
Advait Pillarisetti, Auckland Grammar School
Kacy Zhou, Epsom Girls’ Grammar School
Andrew (Yian) Chen, St Kentigern College
Hyeonmin Seo, Auckland Grammar School
John Voss, Hillcrest High School
Daniel Mar, St Kentigern College
Yong Whan Shin, Auckland Grammar School
Zach Macleod, Macleans College
Joshua Ng, St Kentigern College
Daniel Blaikie, Bayfield High School
Angus Stewart, Macleans College
Matthew Yip, St Patrick’s College (Town)
Robin Wan, Burnside High School
Owen Wang, Macleans College
Matthew Brittain, Tauranga Boys’ College
Owen Wong, Burnside High School
Yang Fan Yun, Macleans College
Ben Chungsuvanich, Tihoi Venture School, St Paul’s Collegiate
Edward Zhou, Burnside High School
Emma Zhao, Macleans College
Dae Young Han, Westlake Boys’ High School
Josh Hogan, Cambridge High School
Shine Wu, Newlands College
Jonathon Lee, Westlake Boys’ High School
First
Second
Third
Year 11 (Form 5) Prize Winners
David Huang
Papanui High School
Samuel Chen
Pakuranga College
Jack Craig
Otago Boys’ High School
Top 30 (in School Order):
Kim Fowler, ACG Strathallan College
Charles Zha, Cambridge High School
Leo Shin, Riccarton High School
Milidu Ratnayake, Auckland Grammar School
Cameron Stevenson, Christ’s College
Andrew Tang, Scots College
Ryojin Ayers, Auckland International College
Andrey Borro, Glendowie College
Seungjun Bang, St Andrew’s College
Yuhe (Daniel) Bai, Auckland International College
Bruce Zhang, Kristin School
Jessy Zhao, St Cuthbert’s College
Seokjoon Hong, Auckland International College
Richard Tang, Macleans College
Chan (Chris) Lee, St Kentigern College
Ningyuan (Michael) Li, Auckland International College
Teresa Zhou, Macleans College
Alec van Helsdingen, St Peter’s College (Epsom)
Brian Qi, Auckland International College
Hamish Weir, Onslow College
Matthew Beardsworth, Te Aho o Te Kura Pounamu
Zhifei (Tony) Shen, Auckland International College
Edward Chen, Palmerston North Boys’ H.S.
Su A Kim, Waikato Diocesan School for Girls
Yilun Wu, Auckland International College
Finlay McRae, Palmerston North Boys’ H.S.
Winston Yao, Westlake Boys’ High School
As usual, in most cases only one solution method is given. It is not necessarily the shortest, nor the most elegant.
Question 1 (Year 9 and below only)
Imagine that you earn $1 per minute before tax, and you work for exactly 40 hours per week (eight hours work
per day for 5 days), for exactly 50 weeks a year.
(a)
There are 480 minutes in an eight hour working day. Show (in one line) the calculation needed
to work out this value. 60 × 8. Well answered. Some students gave 480 × 1, which proves nothing.
(b)
How much do you earn (before tax) in one (5 day) week? $2400. Almost everyone answered this.
(c)
How many full days will you need before you earn more than $1000 (before tax) for the first time?
$480 × 2 = $960 (not big enough); $480 × 3 = $1440. Therefore 3 full days needed. Well answered,
although some students didn't show both calculations.
(d)
Show that in one year you will earn exactly $120 000 before tax. 2400 × 50. Some students didn't read
the '50 weeks' at the start of the question. They tried every possible method using '52'.
(e)
If you are taxed at 33% (33 cents for every dollar you earn), how much do you earn in a year after tax?
$80 400. There were numerous arithmetic mistakes here. '1 – 0.33 = 0.77' was common, as was
thinking that 0.33 = 1/3. Some students ended with answers bigger than $120 000.
(f)
How long will it take you to earn a billion dollars ($1000 million or $1 000 000 000) before tax? (Answer
to the nearest year.) 1 000 000 000/120 000 = 8333 years (nearest year). This question was designed to
show how large a billion is. However, many students only worked it out for a million (giving 8 or 9
years). One student estimated you could earn a billion dollars during the second week.
Question 2 (All Years)
The five digits ‘1’, ‘2’, ‘3’, ‘4’, and ‘5’ are combined together to form a five digit number (for example 34251).
(a)
If every digit is allowed to be repeated up to (and including) five times, what is the smallest possible
number created? 11111. Well answered, although many students did not read the words 'five digit
number' at the top and wrote '1111122222333334444455555' as their answer for no credit.
(b)
If no digit is allowed to be repeated, what is the smallest possible number created? 12345. Well
answered by almost everyone.
(c)
If the digit ‘5’ is allowed to be repeated up to (and including) five times, but no other digit is allowed to
repeat, what is the largest possible number created? 55555. Fairly well answered, although the 'five
digit' restriction was again missed by many, with the common wrong answer being '555554321'.
(d)
Assume that every digit is allowed to be repeated up to (and including) five times.
(i)
How many possibilities are there for the first digit? 5. Well answered.
(ii)
Altogether for all five digits there are 3125 possible numbers of this type. Briefly (in no more
than one line) write down the calculation needed to work out this number. 5 × 5 × 5 × 5 × 5 or
55 (either). Fairly well answered. However, we did not give credit to those students who got
their calculators out and found that 625 × 5 = 3125.
(e)
If no digit is allowed to be repeated, how many such numbers are possible? (You do not have to list them
all, but briefly explain your answer.) 5! = 120. There are 5 possible digits for the first 'space', then 4
etc. Answered by about half the candidates.
(f)
If the digit ‘5’ is allowed to be repeated up to (and including) five times, but no other digit is allowed to
repeat, how many such numbers are possible? (You do not have to list them all, but briefly explain your
answer.)
There can be all numbers being 5 (i.e. 55555). There can be four numbers being 5, with the other
number fitting in anywhere, e.g.15555, 55255 etc. There can be three 5s with two other numbers
e.g. 55352 etc., and so on. Each type has to be worked out separately, then they are added up.
(One student did it correctly with full working in one line.)
all five 5's present: 1 possibility
four 5's present: 5 × 4 = 20 possibilities
three 5's present: 10 × 12 = 120 possibilities
two 5's present: 10 × 24 = 240 possibilities
one 5 present: 120 possibilities (the same as in (e))
Total: 501.
We did not do a full count, but this was answered correctly by fewer than 30 candidates (out of
nearly 7000).
Question 3 (All Years)
(a)
Write 1/2 + 1/3 as a single fraction in simplest form. 5/6. Well answered, although the 'expected'
wrong answer of '2/5' occurred, especially but not exclusively among Year 9 students.
(b)
Every whole number or fraction can be written as the sum of unit fractions (a fraction with a 1 as the
numerator).
For example, 23/24 = 1/2 + 1/4 + 1/8 + 1/12
(i)
Write 11/12 as the sum of any number of unit fractions. 1/2+1/4+1/6 or 1/2+1/3+1/12. We
made a mistake with the question. We should have said 'different unit fractions'. We gave
credit to those students who used repeated fractions. Other answers exist, but were rare.
(ii)
Write 1 as the sum of exactly three different unit fractions. 1/2+1/3+1/6. This, like the previous
answer, was fairly well done, but some students saw the question, thought 'Fractions' and
missed the question completely out.
(iii)
Write 4019/4074 as the sum of four different unit fractions. 1/2+1/3+1/7+1/97. The correct
answer was not uncommon. Some answered this with no working (probably doing it entirely
on their calculator). Others missed it out completely. The clue is to factorise 4074. Since
4074 = 2 × 3 × 7 × 97, the answer follows quickly.
(c)
What is the last digit of 22015? 8. There is a repeating pattern: 21 = 2, 22 = 4, 23 = 8 ,24 = 16, 25 = 32
etc. Since 2015 / 4 = 503.75, it must end in the third number in the sequence. A 'standard' problem. It
was well done in some schools, but unknown in others. If you are doing any problem-solving, this type of
question should be one of the first you look at. It was also noticeable that some students who started with
22 = 4 instead of 21 = 2, used the 'right' method (for partial credit) but reached the wrong answer.
One alternative approach is to note that a number ending in 6 raised to any power will also end in 6.
Now 22015 = 2322012 = 8(24)503 = 8(16)503. Since (16)503 must end in 6 and 8 × 6 = 48, it follows that 22015
must also end in 8.
Question 4 (All years)
Zoey and her older sister Fiona were playing with the new calculators they were given for Christmas. After a
while, Zoey said ‘Cool. If I subtract 72 from 112 I get 72 and if I subtract 32 from 92 I also get 72. I bet there are no
other pairs of squares which subtract to give 72.’
Fiona replied ‘I’m not so sure. I think there could be at least another pair of numbers that work. But we’ll have
to use some algebra to find out.’ She let x and y be any two numbers with x > y (x larger than y).
(a)
Fiona felt that the two expressions (x + y) and (x – y) were important. First she set x to 11 and y to 7,
and found the value of the expression (x + y) + (x – y). What is this value? (11 + 7) + (11 – 7) = 22.
This should have been easy, and for many students it was. But 14 (from those who changed the '+' to a
' –' ) and 72 (from those who changed '+' to '×') were too common.
(b)
She then got Zoey to show that (x + y) + (x – y) is always even, for all natural numbers x and y. How is
this done? (x + y) + (x – y) = 2x. Therefore always even. Sometimes well done.
(c)
Show that (x + y)(x – y) = x2 – y2, for all x and y, the next step in the process. Since (x + y)(x – y) =
x2 – xy + xy – y2, this simplifies to x2 – y2 (given). We were looking for the 'cancellation'. Fairly well
done by Year 11 students who, of course, have studied this, although many of those who answered
this correctly failed to answer the algebraically easier part (b) correctly, which was disappointing.
(d)
Next Fiona showed that (x + y) and (x – y) are either both even or both odd. How did she do this? From
(b), (x + y) + (x – y) is always even. So if (x + y) is even, then so is (x – y). (Similarly, if one is odd, so is the
other.) Not easily answered. Too many gave vague, incomprehensible, answers.
(e)
Finally she investigated whether there were more pairs of numbers x and y which worked in the
equation x2 – y2 = 72 (apart from 11 and 7, and 9 and 3, which Zoey had found). Who was right? (If you
think Zoey was right, prove it by showing there are no more pairs. If you think Fiona was right, find
another pair of numbers which satisfy the equation x2 – y2 = 72.)
72 = x2 – y2 = (x + y)(x – y). Since 72 is even, then both (x + y) and (x – y) must be even.
Pairs of factors of 72 are (1, 72); (2, 36); (3, 24); (4, 16); (6, 12); (8, 9).
Only the 'even pairs' (in bold) yield solutions.
Solution 1: x + y = 36; x – y = 2. Solve simultaneously to get 2x = 38. So x = 19, y = 17.
(Solution 2: x + y = 16; x – y = 4 yields x2 = 11, y2 = 7 (already found).
Solution 3: x + y = 12; x – y = 6 yields x3 = 9, y3 = 3) (already found).
So Fiona was right. The 'extra' pair is x = 19, y = 17.
Many students found this answer (from their calculator?) without working (only partial credit). Not
badly answered by some Year 11 students.
Question 5 (All Years)
Three snooker balls, each of radius 2.5 cm, are
placed side by side on a snooker table so that they
are just touching each other (see Illustration 1).
(a)
(b)
Describe exactly the shape formed when the
centres A, B, and C of the snooker balls are
joined up.
Equilateral triangle. Quite a few students
missed the word 'equilateral' out.
Show that the distance from A to M, where M is
the midpoint of the side BC, is exactly (5/2)√3 cm.
(If you cannot obtain this value, use it in part (d).
The sides of the triangle are 5 cm.
The height of the triangle can be found using
Pythagoras:
h2 = 52 – (5/2)2.
h2 = 25(1 – 1/4)
h2 = 25 × 3/4
h = 5/2√3 (given)
Not badly done by many Year 11 students.
Illustration 1
A fourth snooker ball, with exactly the same radius, is
placed carefully on top of the other three balls so that it
just touches each of the other three balls (see Illustration
2). We will call the centre of this top ball D.
(c)
With one word (beginning with ‘t’), name the
shape formed when the centres of the four balls
(A, B, C, and D) are joined up. Tetrahedron. Well
answered although 'trapezium' was common.
Illustration 2
(d)
Find the height from the top (T) of the fourth snooker ball to the table.
Let the altitude of the tetrahedron be a. Other points and distances are shown in the diagram.
From Y to M must be (5√3/2 – y).
We have two simultaneous equations to solve:
(1)
a2 + y2 = 25.
(2)
a2 + (y – (5/2)√3))2 = ((5/2)√3))2.
We eventually get y = 5/√3.
Substituting this into (1) gives a = (5√6)/3.
So the height from T to table is
(5√6)/3 + 5 = 9.082482905 cm
= 9.1 cm (1 d.p.)
The hardest question in the competition, only
answered by a few.
(e)
A snooker table measuring (length) 355 cm by (width)
180 cm, with (height) 85 cm is in a room with a single
light source directly above the middle of the table. The shadow of the width on the floor is measured at
260 cm (see Illustration 3). How high is the light above the top of the table?
The pyramids from the light to the table and from the light to the floor are similar.
Therefore
h/180 = (h + 85)/260
260h = 180(h + 85)
260h – 180h = 15 300
80h = 15 300
h = 191.25 cm
The enlargement was solved by only a few students, although many gave the scale factor correctly. It was
noticeable that this was the only question the Year 10 winner got wrong. We think he ran out of time. The
Year 11 winner got the whole competition correct, even giving alternative solution methods in places.
Illustration 3
Hints
1. Don't give more than one answer. The markers work on the principle: 'Two answers. One
wrong. Zero credit'. This happens even if one of the answers is correct.
2. Show necessary working. For example, in this year's competition there was a question
where you had to work out the value of (11 + 4) + (11 – 4). The correct answer was 22.
However, several students reached the wrong answer 14. Some of them gave the answer
only (14) and they earned 0. However, if you wrote '(15) + (7) = 14', you at least got half
credit, even though the answer is wrong.
Another example was showing if Fiona or Zoey was right. Many students worked out that
Fiona was right, because 192 – 172 = 72. However, the answer only earns one fifth of the
total credit for the question. You had to show working to do well in the question.
3. Don't write English sentences as your answer unless you're asked for one. 'Show necessary
working' almost certainly means 'show mathematical working'. This year, one student spent
three lines explaining why there are sixty minutes in an hour. We think that we already knew
that. All we wanted to see was '8 × 60 (= 480)' in the question. The words 'times' and 'equals'
waste time when there are perfectly good symbols ('×' and '=') available.
In 4(b), what we were looking for was '(x + y) + (x – y) = 2x. Always even.' English
sentences gained little, if any, credit.
Another student wrote 10 line answers for questions 1(a) to 1(d), taking a whole page. We
are sure she later wondered why she ran out of time.
4. It is a good idea to write useful information in your own writing at the start of each
question. For example in Question 1 this year you were told '50 weeks'. Some students
wrote '50 weeks' down at the top, but others didn't read it properly. They used '52 weeks' and
later wondered why they couldn't reach the given answers.
5. Don't forget to answer questions in order. Those Year 10 and Year 11 students who started
with Questions 4 or 5 almost certainly guaranteed themselves a poor mark. One Year 11
student spent three pages answering Question 4 first (for little credit), then ran out of time to
answer much else. Her total score was less than 10.
6. Too many students don't write down what's actually there. For example, in Question 4(a)
many students missed out the '+' or changed it to a '–'. Be careful.
7. Don't use technical drawing pencils. They are hard to read, and if you don't get full credit
because it is too hard to read an answer, we're afraid it's your own fault.
8. Have a go! We do not give negative marks for wrong answers. If you put something down,
you have a better chance than if you leave the answer blank.
9. Don't forget to write your name at the top. This year, about five students forgot to do so,
and two of them could not be identified by their school at all.
Don't forget to visit our web page www.maths.otago.ac.nz/jmc:
There you will find, among other things, questions and model answers from recent years.
We also have a Twitter account at @uojmc.